What have I done wrong?

[kenhung123]

Sketch the graph of f(x)= 2^x under the following transformations

2f(x-1)+1

x'-1=x
x'=x+1

2y'+1=y
y'=(y-1)/2

(x,y)=>(x+1,(y-1)/2)

(1,1/2)=>(2,-1/4)
(0,1)=>(1,0)
(1,2)=>(2,1/2)

[the.watchman]

I'd suggest you don't apply the 'new-image' method to do this question

So if

Then

[stonecold]

yeah you basically follow bodamas.

so first you multiply the function by 2,
then you sub in whatever is the brackets for x,
then just add the 1 on in the end...

[kenhung123]

Oh thanks guys.
Also, I am wondering in this case the base is 4? Does the base ever change due to transformations and so "a" changes? Or ultimately the transformations balances out and the base always remains original?

[kenhung123]

I'd suggest you don't apply the 'new-image' method to do this question

So if

Then

Hmm ok, so how do you find the image of the 3 points?

[the.watchman]

Oh thanks guys.
Also, I am wondering in this case the base is 4? Does the base ever change due to transformations and so "a" changes? Or ultimately the transformations balances out and the base always remains original?

Not usually, unless you have and

I'd suggest you don't apply the 'new-image' method to do this question

So if

Then

Hmm ok, so how do you find the image of the 3 points?

Sub the x-values into both the old and new functions

[TrueTears]

sub them into 2 \cdot 2^{x-1} + 1

[m@tty]

You can simplify the function:



So the dilation by 2 in y has cancelled the translation in x.

Hence you only have to add 1 unit to the y component of each point from .

[the.watchman]

You can simplify the function:



So the dilation by 2 in y has cancelled the translation in x.

Hence you only have to add 1 unit to the y component of each point from .

Noice
That's really smart, I'd better apply it in future

[m@tty]

It is quite useful sometimes. It simplifies computation dramatically in this instance.

[kenhung123]

You can simplify the function:



So the dilation by 2 in y has cancelled the translation in x.

Hence you only have to add 1 unit to the y component of each point from .

Hmm, I'm not sure why you did {x-1+1) where did the +1 come from?

[TrueTears]

a^b . a^c = a^(b+c)

[kenhung123]

Oh! That is very interesting

[kenhung123]

Oh thanks guys.
Also, I am wondering in this case the base is 4? Does the base ever change due to transformations and so "a" changes? Or ultimately the transformations balances out and the base always remains original?

Not usually, unless you have and

I'd suggest you don't apply the 'new-image' method to do this question

So if

Then

Hmm ok, so how do you find the image of the 3 points?

Sub the x-values into both the old and new functions

Like in this question we are given 2^x, we need to sketch the graph under the transformations. So can we always assume "a"=2? Or we need to find the actual equations and not only transformation and apply it to the 3 points?

[superflya]

Oh! That is very interesting

lol index laws